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An infinitely long rod lies along the axis of a concave mirror of focal length f. The near end of the rod is distance `u gt f` from the mirror. Its image will have lengthA. `f^(2)/(u-f)`B. `(uf)/(u-f)`C. `f^(2)/(u+f)`D. `(uf)/(u+f)` |
Answer» Correct Answer - A (a) From the relation, `1/v+1/u=1/for1/v-1/u=1/(-f)` `1/v=1/u-1/forv=((uf)/(u-f))` Since,`ugtf`,v is negative or `abs(v)=((uf)/(u-f))gtf` The end which is at infinity will have its image at focus. `therefore` Length of image, `L=abs(v)-f=f^(2)/(u-f)` |
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